solve the given differential equation subject to initial conditions.

Question

2y"+y'-y=x+1;   y"(0)=1,   y'(0)=0

Answer 1

2y"+y'-y=x+1.

Let y=y_c+y_p, where y_c is the characteristic solution and y_p is the particular solution.

2y_c"+y_c'-y_c=0 can be solved by factorising 2r²+r-1=(2r-1)(r+1) which leads to the characteristic solution:

y_c=Ae^½x+Be^-x, where A and B are constants which can be found through applying the initial conditions.

For y_p we "guess" that y_p=ax+b where a and b are constants found by applying the DE:

y_p'=a, y_p"=0, so 2y_p"+y_p'-y_p=a-ax-b=x+1, so a-b=1 and a=-1, so b=-2. Therefore, y_p=-x-2

y=Ae^½x+Be^-x-x-2.

y'=½Ae^½x-Be^-x-1; y"=¼Ae^½x+Be^-x; y'(0)=0=A/2-B-1, A=2B+2, and y"(0)=1=A/4+B=(B+1)/2+B,

B+1+2B=2, B=⅓, A=8/3.

SOLUTION: y=(8/3)e^½x+⅓e^-x-x-2.

CHECK

y'=(4/3)e^½x-⅓e^-x-1, y"=⅔e^½x+⅓e^-x; y'(0)=4/3-1/3-1=0; y"(0)=2/3+1/3=1;

2y"+y'-y=(4/3)e^½x+⅔e^-x+(4/3)e^½x-⅓e^-x-1-(8/3)e^½x-⅓e^-x+x+2=x+1, so the solution checks out OK.